Why doesn't the join operation on the category of simplicial sets commute up to unique isomorphism? I mean, aren't products and coproducts commutative up to isomorphism? That leads me to conclude at first glance that the join is commutative, but it's not. Recall, given two simplicial sets $S$ and $S'$, we define the join to be the simplicial set such that for all finite nonempty totally ordered sets $J$, $$(S\star S')(J)=\coprod_{J=I\cup I'}S(I) \times S'(I')$$
Where $\forall (i \in I \land i' \in I') i < i'$, which implies that $I$ and $I'$ are disjoint.

Now the thing is, clearly my conclusion is stupid, because we use the fact that it doesn't commute to distinguish between over quasi-categories and under quasi-categories. Where did I go wrong?

I hope this is up to the standards of MO, but if it's not, I'll delete the topic.

3 Answers
3

Implicit in the index of the coproduct is that you're writing J as an ordered disjoint union of I and I', where I comes first.

EDIT: Some elaboration.

For a simplicial set $T$, let's write $T\_n$ for the "n-simplices", i.e. the value of on the ordered set $\{0,1,...,n\}$; these together with the maps between them determine the functor $T$ completely. (Your formula for the join requires the convention that $T$ takes the empty set to a single point.)

Given $S$ and $S'$, let's determine the 0- and 1-simplices of the join.

First, $(S \star S')\_0$. There are exactly two ways to write $\{0\} = I \cup I'$ in an order preserving way as indexed by the coproduct: either $I'$ is empty and $I$ is everything, or vice versa. Thus $(S \star S')\_0 = S\_0 \cup S'\_0$ accordingly. The zero-simplices of the join are the zero-simplices of the original simplicial set.

Next, the 1-simplicies. Similarly
$$
(S \star S')\_1 = S(\{0,1\}) \cup (S(\{0\}) \times S'(\{1\})) \cup S'(\{0,1\})= S\_1 \cup (S\_0 \times S'\_0) \cup S'\_1
$$
There are then 3 types of 1-simplices: the 1-simplices from S, those from S', and for each choice of a point of S and a point of S' there is a new 1-simplex.

The two boundary maps $(S \star S')\_1 \to (S \star S')\_0$ are induced by the inclusions of $\{0\}$ and $\{1\}$ into $\{0,1\}$ (the "back" and "front" boundaries respectively). In particular, on the new 1-simplices $S\_0 \times S'\_0$ the back boundary is the projection to $S\_0$ and the front boundary is projection to $S'\_0$. There is asymmetry here because the only ways we're allowed to decompose $\{0,1\}$ in the coproduct have $I$ (the subset corresponding to $S$) first and $I'$ second. None of the "new" edges start at a vertex of $S'$ and end at a vertex of $S$.

Actually, I don't see how this gives us a direction or order. Could you elaborate a little further, preferably not with degenerate simplicial sets? I don't see what you mean by ordered disjoint union.
–
Harry GindiNov 29 '09 at 19:34

I added some more that will hopefully clarify.
–
Tyler LawsonNov 29 '09 at 22:12

It might be helpful to work through some simple examples. You probably know that Δn ★ Δk = Δn+k+1. This has to do with the ordinal sum: one way of defining joins is as a restriction of the monoidal structure on augmented simplicial sets, which are contravarient functors from the category Δ+ of all finite ordinals (including the empty ordinal) into sets. The category Δ+ has a monoidal structure given by ordinary addition with ∅ as the unit, and this induces the aforementioned monoidal structure on augmented simplicial sets. The thing we call n when we are talking about simplicial sets is really the ordinal n+1, so the formula above holds because

(n+1) + (k+1) = (n+k+1)+1.

Of course, this example doesn't illustrate the asymmetry you asked about, but this one will:

I'm not familiar with the notation $\Lambda^k [j]$. I am familiar with $\Lambda^k_j$ being the jth k-horn, but $0\leq j \leq k$ in that case.
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Harry GindiNov 30 '09 at 10:25

Sorry. Both notations are fairly standard. I chose the former because it looked better in html. But when writing $\Lambda^k[j]$, one typically means for $j$ to be the dimension and $0 \leq k \leq j$ to indicate which face is missing. (So this is $\Lambda^j_k$ in your notation.)
–
Emily RiehlDec 1 '09 at 5:19

As mentioned in the other answers, the join of simplicial sets is closely related to "ordered disjoint union", or "concatenation", of (totally, partially, pre-) ordered sets. You can use this both to get simple examples of its non-commutativity, and to help reconcile that with the intuition that it should be commutative.

Any order $X$ can be seen as the simplicial set whose $n$-simplices are chains $(x_0 \leq \ldots \leq x_n)$ from $X$. That is, there's a full and faithful "nerve" embedding $N: \mathrm{PreOrd} \rightarrow \mathrm{SSet}$. Now if $X$ and $Y$ are orders, seen as their nerves, $X \star Y$ is exactly (the nreve of) their ordered disjoint union.

So e.g. $1 \star \mathbb{N} \not \cong \mathbb{N} \star 1$ is an easy and intuitive example of the non-commutativity.

On the other hand, like you say, looking at the definition, there is an immediate intuition that it should be commutative in some sense, and chasing it down, I think what that intuition is coming from is something like the fact: for any simplicial sets $X$, $Y$,

$(X \star Y)^\mathrm{op} \cong Y^\mathrm{op} \star X^\mathrm{op}.$

So commuting $\star$ distributes over $\mathrm{op}$: so in a sense, the only asymmetry in $\star$ is an asymmetry of variance. This is nice and intuitive for ordered sets, and easily shown for all simplicial sets.